Question

Two integers $x$ and $y$ are chosen with replacement out of the set $0,1,2,3,...,10$ . Then the probability that $|x-y|>5$ is

• A. $\frac{81}{121}$
• B. $\frac{30}{121}$
• C. $\frac{25}{121}$
• D. $\frac{20}{121}$

Answer 1

The correct option is B $\frac{30}{121}$

Possible combinations of (x, y) are:
$(0,6),(0,7),(0,8),(0,9),(0,10),(1,7)...(4,10)$ and their reverse.
Thus, the number of combinations are = $(5+4+3+2+1)\ast 2=30$
Total no. of combinations (with replacement, hence numbers may be repeated) = $11\ast 11=121$
Therefore, probability = $\frac{30}{121}$